# Maps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum

Let $$ku$$ be the connective cover of the complex $$K$$-theory spectrum $$KU$$. Consider the mod-$$p$$ Eilenberg-MacLane spectrum $$H\mathbb{Z}/p$$.

I want to see that $$[H\mathbb{Z}/p,ku]=0$$.

Since $$H\mathbb{Z}/p$$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories") the result will follow once we know that $$ku$$ is harmonic (Corollary 4.11 in loc. cit.)

Is it true that the spectrum $$ku$$ is harmonic?

If not, how can I prove the claim?

Certainly, an idea is to use directly the theorem of Margolis (“Eilenberg-MacLane Spectra”): For any spectrum $$Y$$ of finite type, there is an isomorphism $$[H\mathbb{Z}/p,Y]\to \text{Hom}_{\mathcal{A}}(H^{*}(Y,\mathbb{Z}/p),\mathcal{A})$$ where $$\mathcal{A}$$ is the Steenrod algebra.

Is it true that $$\text{Hom}_{\mathcal{A}}(H^{*}(ku,\mathbb{Z}/p),\mathcal{A})=0$$?

If the answer is yes, why is that?

You can also cheat: since HZ/p is connective,$$[HZ/p,ku] = [HZ/p,KU] = [HZ/p \wedge KU,KU]_{KU-modules} = 0$$ since $$HZ/p \wedge KU = 0$$.

Edit: Let me stress that the other answers give more information, namely the calculation of maps from HZ/p to ku in all degrees.

• Dear @DustinClausen. I can't see why is it true that $H\mathbb{Z}/p$ smash with $KU$ is 0. – user438991 Apr 15 '19 at 6:52
• Dear user438991, I think this fact has come up a few times on MathOverflow. Anyway, here are two proofs. One is using the theory of complex oriented cohomology theories: HZ/p gives the additive formal group and KU gives the multiplicative formal group, so on HZ/p smash KU you must have a formal group which is isomorphic to both the additive and multiplicative one. But this is impossible over a ring of characteristic p, because the heights are different. – Dustin Clausen Apr 15 '19 at 8:23
• Another proof is using Adams' self-map v_1: \Sigma^d S/p --> S/p. You can pin down the minimal d (2p-2 for odd p, 8 for p=2), but the only important thing here is that it's positive. When you smash with KU this v_1 becomes a power of the Bott element and hence is invertible, but when you smash with HZ it becomes null for degree reasons. Hence HZ/p smash KU ( = S/p smash HZ smash KU) carries a self-map which is both invertible and null, therefore it must vanish. – Dustin Clausen Apr 15 '19 at 8:26
• I guess there are also more computational proofs. You could brute force calculate HZ/p smash KU: by Bott perodicity, it is (with shifts) a certain N-indexed colimit of Z/p-homologies of the space BU ( = colim n BU(n) ). The latter is a well-known calculation, and then it's a matter of identifying the effect of the Bott map \Sigma^2 BU --> BU on homology to be able to pass to the colimit. I guess the key for the vanishing should be the result (of Bott) that the Hurewicz image of the n^{th} power Bott element in \pi_{2n}BU is divisible by (n-1)!, hence dies (mod p) for large n. – Dustin Clausen Apr 15 '19 at 10:11
• You can probably shortcut the computational proof by using Snaith's presentation KU = \Sigma^\infty_+ BU(1)[bott^{-1}]. Then it's enough to invert Bott on the homology of the space BU(1). That this dies with finite coefficients follows from the fact that the pontryagin ring structure on H_*(BU(1)) is a divided power algebra. – Dustin Clausen Apr 15 '19 at 10:12

There are cofibrations $$\Sigma^2 kU\xrightarrow{v} kU\to H\xrightarrow{\alpha}\Sigma^3 kU$$ and $$H\xrightarrow{p}H\to H/p\xrightarrow{\beta} \Sigma H.$$ The composite $$\alpha\beta\colon H/p\to\Sigma^4 kU$$ is nontrivial. In fact, I am fairly sure that the map $$(\alpha\beta)^*$$ induces an isomorphism $$(kU^*kU)/(p,v)\to kU^*(H/p)$$. To prove this (using the above cofibrations), one would only need to check that the sequence $$(v,p)$$ is regular on $$kU^*kU$$, and I think that that is known.

We get essentially the same picture if we replace $$kU$$ by the Adams summand $$BP\langle 1\rangle$$ as in Ekie's answer. Then you can use the Adams spectral sequence which starts with $$\text{Ext}^{**}_{E(1)}(\mathbb{F}_p,\mathcal{A})$$. Here $$\mathcal{A}$$ has the form $$E(1)\otimes R$$ for some $$R$$, and $$E(1)$$ is self-dual up to a shift so we just get $$\text{Ext}^{**}=\text{Ext}^{0*}=\Sigma^d R$$, where $$d=|v_0|+1+|v_1|+1=2p$$.

• Ha. We were apparently posting basically the same thing at the same time. – Nicholas Kuhn Apr 11 '19 at 16:47
• @NeilStrickland: I am reading your answer and I feel I have a question, which may not directly be related to the above. I just wondering what is $[H\underline{\mathbb{Z}/2}, \Sigma H\underline{\mathbb{Z}/2}]^{C_2}$? – Surojit Ghosh Jul 24 at 9:02

Lets just work with $$p=2$$. $$ku$$ is certainly not harmonic, but $$[H\mathbb Z/2, ku]=0$$. The Adams spectral sequence works very nicely to show all of this: The $$E_2$$--term is $$Ext^{s,t}_{A}(H^*(ku), H^*(H\mathbb Z/2)) = Ext^{s,t}_{A}(A//E(1), A) = Ext^{s,t}_{E(1)}(\mathbb Z/2, A)$$. Since $$A$$ is a free module over the Hopf algebra $$E(1)$$, and we conclude that $$A$$ is an injective $$E(1)$$-module. Thus the Adams $$E_2$$--term is just $$Hom$$, and we learn that there is an isomorphism $$[H\mathbb Z/2, \Sigma^tku] = Hom_{E(1)}(\Sigma^t\mathbb Z/2, A)$$. The smallest $$t$$ for which this is nonzero is $$t=4$$, corresponding to the element $$Sq^1Sq^2Sq^1 \in A$$. The associated map $$H\mathbb Z/2 \rightarrow \Sigma^4 ku$$ can be taken to be the composite of the `Bockstein' map $$H\mathbb Z/2 \rightarrow \Sigma H\mathbb Z$$ with one suspension of the map $$H\mathbb Z \rightarrow \Sigma^3 ku$$ arising from the cofibration sequence $$\Sigma^2 ku \rightarrow ku \rightarrow H\mathbb Z \rightarrow \Sigma^3 ku.$$

I am sure the odd prime version is similar.

• Is ku really not harmonic? Here's a back-of-the-envelope calculation: chromatic fracture presents $L_1 ku$ as an extension of $ku$ by $Q / Z \otimes \Sigma^{-1} KU(-\infty, 0)$. $L_{K(< \infty)} ku := L_{\bigvee_{j < \infty} K(j)} ku$ is $L_1 ku$ again, so the harmonic localization $L_{K(\le \infty)} ku$ can be recovered from a fracture square involving $L_1 ku$ and $L_{K(\infty)} ku$. $L_{K(\infty)} ku$ is $ku^\wedge_p$, so $L_{K(< \infty)} L_{K(\infty)} ku$ has a similar coconnective part, and I think the coconnective parts cancel in the fracture square to give $L_{K(\le \infty)} ku = ku$. – Eric Peterson Apr 16 '19 at 13:46
• I imagine I'm missing something, not you, but I'd of course like to know what! – Eric Peterson Apr 16 '19 at 13:47
• @Eric Peterson The fact that there are nontrivial maps from the "dissonant" spectrum HZ/2 to a suspension of ku proves that ku is not harmonic. – Nicholas Kuhn Apr 16 '19 at 19:07
• Oh: harmonic localization is $L_{K(< \infty)}$, not $L_{K(\le \infty)}$, or else HZ/2 itself would be recoverable. Great, thanks! – Eric Peterson Apr 16 '19 at 19:11

(Edited to include the case when $$p$$ is odd. Hopefully there are no mistakes as I don't usually work with $$p$$ odd.)

This is true, assuming you mean degree 0 maps. We have $$ku\simeq l \vee \Sigma^{2} l \vee \dots \vee \Sigma^{2(p-2)} l$$ where $$l$$ is called the Adams summand (usually denoted $$BP\langle 1 \rangle$$). Notice when $$p=2$$ that $$ku=l$$. The cohomology of $$l$$ is $$H^*(l; \mathbb{F}_p) \cong \mathcal{A}//E(1) = \mathcal{A} \otimes_{E(1)} \mathbb{F}_p$$ where $$E(1)$$ is the exterior algebra generated by the Milnor primitives $$Q_0$$ and $$Q_1$$ in degrees 1 and $$2p-1$$, respectively. For example, if $$p=2$$, then $$Q_0 = Sq^1$$ and $$Q_1 = Sq^2 Sq^1 + Sq^3$$. Thus $$H^*(ku) \cong \mathcal{A}//E(1) \oplus \Sigma^2 \mathcal{A}//E(1) \oplus \dots \oplus \Sigma^{2p-4} \mathcal{A}//E(1).$$

Now look at each factor: by change of rings, $$\text{Hom}_{\mathcal{A}}(\Sigma^{2n} H^{*}(l;\mathbb{F}_p),\mathcal{A}) \cong \text{Hom}_{E(1)}(\Sigma^{2n} \mathbb{F}_p, \mathcal{A}|_{E(1)})$$ for $$0\leq 2n \leq 2p-4$$, where $$\mathcal{A}|_{E(1)}$$ means restriction to $$E(1)$$.

There are no nonzero $$E(1)$$-maps from $$\mathbb{F}_p$$ to $$\mathcal{A}|_{E(1)}$$ because the bottom class of $$\mathcal{A}$$ supports nontrivial operations (e.g. $$Q_0$$). This shows the group is 0 for $$n=0$$, which is the only case to consider if $$p=2$$. If $$p$$ is odd, then the odd primary Steenrod algebra is generated by the Bockstein $$\beta$$ in degree 1 and the reduced powers $$P^i$$ in degree $$2i(p-1)$$. The first $$P^i$$ is in degree $$2p-2 > 2p-4$$, hence the generator of $$\Sigma^{2n} \mathbb{F}_2$$ can never hit a nonzero class when $$n>0$$, so $$\text{Hom}_{E(1)}(\Sigma^{2n} \mathbb{F}_p, \mathcal{A}|_{E(1)}) = 0$$ for all $$n$$ with $$0\leq 2n \leq 2p-4$$.

As noted in other answers, this statement is not true if we look at suspensions of $$ku$$ instead, because e.g. there is a nontrivial map $$\Sigma^4 \mathbb{F}_2 \to \mathcal{A}$$ where the generator of $$\Sigma^4 \mathbb{F}_2$$ hits $$Q_0Q_1$$ (since $$Q_0Q_1$$ supports no operation when $$\mathcal{A}$$ is restricted to $$E(1)$$, there is no longer a contradiction here).

• Dear @Ekie. Are you proving the claim in the OP question for $ku_{(p)}$? I'm a little confused. – Tsk Apr 15 '19 at 6:49
• @Tsk I am proving the claim that $Hom_{A}(H^*(ku; \mathbb{Z}/p), A) = 0$, for normal $ku$. I can understand if my answer seems convoluted, as it has grown/changed a little bit. Let me know which parts are confusing, and I can try to address that. – Ekie Apr 16 '19 at 1:39
• Ah, I see, yes I'm definitely using a $p$-local splitting. Hmm I feel as though the module argument still works out when using $p$-completion instead, which would give the right answer for mod $p$ homology, but I'm not sure. In any case, this works for regular $ku$ when $p=2$. – Ekie Apr 17 '19 at 22:53